Let ${(A, \mathfrak{m})}$ be a regular local (noetherian) ring of dimension ${d}$. In the previous post, we described loosely the local cohomology functors

$\displaystyle H^i_{\mathfrak{m}}: \mathrm{Mod}(A) \rightarrow \mathrm{Mod}(A)$

(in fact, described them in three different ways), and proved a fundamental duality theorem

$\displaystyle H^i_{\mathfrak{m}}(M) \simeq \mathrm{Ext}^{d-i}(M, A)^{\vee}.$

Here ${\vee}$ is the Matlis duality functor ${\hom(\cdot, Q)}$, for ${Q}$ an injective envelope of the residue field ${A/\mathfrak{m}}$. This was stated initially as a result in the derived category, but we are going to use the above form.

The duality can be rewritten in a manner analogous to Serre duality. We have that ${H^d_{\mathfrak{m}}(A) \simeq Q}$ (in fact, this could be taken as a definition of ${Q}$). For any ${M}$, there is a Yoneda pairing

$\displaystyle H^i_{\mathfrak{m}}(M) \times \mathrm{Ext}^{d-i}(M, A) \rightarrow H^d_{\mathfrak{m}}(A) \simeq Q,$

and the local duality theorem states that it is a perfect pairing.

Example 1 Let ${k}$ be an algebraically closed field, and suppose that ${(A, \mathfrak{m})}$ is the local ring of a closed point ${p}$ on a smooth ${k}$-variety ${X}$. Then we can take for ${Q}$ the module

$\displaystyle \hom^{\mathrm{top}}_k(A, k) = \varinjlim \hom_k(A/\mathfrak{m}^i, k):$

in other words, the module of ${k}$-linear distributions (supported at that point). To see this, note that ${\hom_k(\cdot, k)}$ defines a duality functor on the category ${\mathrm{Mod}_{\mathrm{sm}}(A)}$ of finite length ${A}$-modules, and any such duality functor is unique. The associated representing object for this duality functor is precisely ${\hom^{\mathrm{top}}_k(A, k)}$.

In this case, we can think intuitively of ${H^i_{\mathfrak{m}}(A)}$ as the cohomology

$\displaystyle H^i(X, X \setminus \left\{p\right\}).$

These can be represented by meromorphic ${d}$-forms defined near ${p}$; any such ${\omega}$ defines a distribution by sending a function ${f}$ defined near ${p}$ to ${\mathrm{Res}_p(f \omega)}$. I’m not sure to what extent one can write an actual comparison theorem with the complex case. (more…)

So I’ve missed a few days of MaBloWriMo. But I do have a talk topic now (I was mistaken–it’s actually tomorrow)! I’ll be speaking about some applications of Sperner’s lemma. Notes will be up soon.

Today I want to talk about how depth (an “arithmetic” invariant) compares to dimension (a “geometric” invariant). It turns out that the geometric invariant wins out in size. When they turn out to be equal, then the relevant object is called Cohen-Macaulay. This is a condition I’d like to say more about in future posts.

0.5. Depth and dimension

Consider an ${R}$-module ${M}$, which is always assumed to be finitely generated. Let ${I \subset R}$ be an ideal with ${IM \neq M}$. We know that if ${x \in I}$ is a nonzerodivisor on ${M}$, then ${x}$ is part of a maximal ${M}$-sequence in ${I}$, which has length ${\mathrm{depth}_I M}$ necessarily. It follows that ${M/xM}$ has a ${M}$-sequence of length ${\mathrm{depth}_I M - 1}$ (because the initial ${x}$ is thrown out) which can be extended no further. In particular, we find

Proposition 15 Hypotheses as above, let ${x \in I}$ be a nonzerodivisor on ${M}$. Then$\displaystyle \mathrm{depth}_I (M/xM) = \mathrm{depth} M - 1.$

This is strikingly analogous to the dimension of the module ${M}$. Recall that ${\dim M}$ is defined to be the Krull dimension of the topological space ${\mathrm{Supp} M = V( \mathrm{Ann} M)}$ for ${\mathrm{Ann} M}$ the annihilator of ${M}$. But the “generic points” of the topological space ${V(\mathrm{Ann} M)}$, or the smallest primes in ${\mathrm{Supp} M}$, are precisely the associated primes of ${M}$. So if ${x}$ is a nonzerodivisor on ${M}$, we have that ${x}$ is not contained in any associated primes of ${M}$, so that ${\mathrm{Supp}(M/xM)}$ must have smaller dimension than ${\mathrm{Supp} M}$. That is,

$\displaystyle \dim M/xM \leq \dim M - 1.$ (more…)

Thanks to all who responded to the bleg yesterday. I’m still haven’t completely decided on the topic owing to lack of time (I actually wrote this post last weekend), but the suggestions are interesting. My current plan is, following Omar’s comment, to look at Proofs from the Book tomorrow and pick something combinatorial or discrete-ish, like the marriage problem or Arrow’s theorem. I think this will be in the appropriate spirit and will make for a good one-hour talk.

4. ${\mathrm{Ext}}$ and depth

One of the first really nontrivial facts we need to prove is that the lengths of maximal ${M}$-sequences are all the same. This is a highly useful fact, and we shall constantly use it in arguments (we already have, actually). More precisely, let ${I \subset R}$ be an ideal, and ${M}$ a finitely generated module. Assume ${R}$ is noetherian.

Theorem 12 Suppose ${M}$ is a f.g. ${R}$-module and ${IM \neq M}$. All maximal ${M}$-sequences in ${I}$ have the same length. This length is the smallest value of ${r}$ such that ${\mathrm{Ext}^r(R/I, M) \neq 0}$.

I don’t really have time to define the ${\mathrm{Ext}}$ functors in any detail here beyond the fact that they are the derived functors of ${\hom}$. So for instance, ${\mathrm{Ext}(P, M)=0}$ if ${P}$ is projective, and ${\mathrm{Ext}(N, Q) = 0}$ if ${Q}$ is injective. These ${\mathrm{Ext}}$ functors can be defined in any abelian category, and measure the “extensions” in a certain technical sense (irrelevant for the present discussion).

So the goal is to prove this theorem. In the first case, let us suppose ${r = 0}$, that is there is a nontrivial ${R/I \rightarrow M}$. The image of this must be annihilated by ${I}$. Thus no element in ${I}$ can act as a zerodivisor on ${M}$. So when ${r = 0}$, there are no ${M}$-sequences (except the “empty” one of length zero).

Conversely, if all ${M}$-sequences are of length zero, then no element of ${I}$ can act as a nonzerodivisor on ${M}$. It follows that each ${x \in I}$ is contained in an associated prime of ${M}$, and hence by the prime avoidance lemma, that ${I}$ itself is contained in an associated prime ${\mathfrak{p}}$ of ${M}$. This prime avoidance argument will crop up quite frequently.

(I don’t know why these <br >’s appear below. If anyone with better HTML knowledge than I could explain what I’m doing wrong, I’d appreciate it!)

Today, we will continue with our goal of understanding some aspects of commutative algebra, by defining depth.

0.3. Depth

Constructing regular sequences sequences is a useful task. We often want to ask how long we can make them subject to some constraint. For instance,

Definition 8
Suppose ${I}$ is an ideal such that ${IM \neq M}$. Then we define the ${I}$-depth of ${M}$ to be the maximum length of a maximal ${M}$-sequence contained in ${I}$. When ${R}$ is a local ring and ${I}$ the maximal ideal, then that number is simply called the depth of ${M}$.

The depth of a proper ideal ${I \subset R}$ is its depth on ${R}$.

The definition is slightly awkward, but it turns out that all maximal ${M}$-sequences in ${I}$ have the same length. So we can use any of them to compute the depth.

The first thing we can prove using the above machinery is that depth is really a “geometric” invariant, in that it depends only on the radical of ${I}$. (more…)